For a Pythagorean triplet built from 2$m,m^2-1,m^2+1$, take $m=4$. What is the largest member of the triplet?

**Step 1**: factorise the number into primes.
**Step 2**: pair up identical factors.
**Step 3**: from each pair, take ONE factor — the product is the square root.

Example: $\sqrt{324}$. Factorise: $324=2^2\times 3^4=(2\times 3\times 3{)}^2$. So $\sqrt{324}=2\times 3\times 3=18$.

Example: $\sqrt{576}$. Factorise: $576=2^6\times 3^2=(2^3\times 3{)}^2$. So $\sqrt{576}=8\times 3=24$.

If any prime appears an odd number of times, the number isn't a perfect square — and you'd need to multiply (or divide) by that prime to make it one.